Query to get only numbers from a string

Question

I have data like this:

string 1: 003Preliminary Examination Plan   
string 2: Coordination005  
string 3: Balance1000sheet

The output I expect is

string 1: 003
string 2: 005
string 3: 1000

And I want to implement it in SQL.

Answer 1

First create this UDF

CREATE FUNCTION dbo.udf_GetNumeric
(
  @strAlphaNumeric VARCHAR(256)
)
RETURNS VARCHAR(256)
AS
BEGIN
  DECLARE @intAlpha INT
  SET @intAlpha = PATINDEX('%[^0-9]%', @strAlphaNumeric)
  BEGIN
    WHILE @intAlpha > 0
    BEGIN
      SET @strAlphaNumeric = STUFF(@strAlphaNumeric, @intAlpha, 1, '' )
      SET @intAlpha = PATINDEX('%[^0-9]%', @strAlphaNumeric )
    END
  END
  RETURN LEN(COALESCE(TRIM(CAST(ISNULL(@strAlphaNumeric, 0) AS INT)),0))>0 then COALESCE(TRIM(CAST(ISNULL(@strAlphaNumeric, 0) AS INT)),0) else 0 end
END
GO

Now use the function as

SELECT dbo.udf_GetNumeric(column_name) 
from table_name

SQL FIDDLE

I hope this solved your problem.

Reference

4/10/23 - Modified Return Statement based on comments

Answer 2

Try this one -

Query:

DECLARE @temp TABLE
(
      string NVARCHAR(50)
)

INSERT INTO @temp (string)
VALUES 
    ('003Preliminary Examination Plan'),
    ('Coordination005'),
    ('Balance1000sheet')

SELECT LEFT(subsrt, PATINDEX('%[^0-9]%', subsrt + 't') - 1) 
FROM (
    SELECT subsrt = SUBSTRING(string, pos, LEN(string))
    FROM (
        SELECT string, pos = PATINDEX('%[0-9]%', string)
        FROM @temp
    ) d
) t

Output:

----------
003
005
1000

Answer 3

Query:

DECLARE @temp TABLE
(
    string NVARCHAR(50)
)

INSERT INTO @temp (string)
VALUES 
    ('003Preliminary Examination Plan'),
    ('Coordination005'),
    ('Balance1000sheet')

SELECT SUBSTRING(string, PATINDEX('%[0-9]%', string), PATINDEX('%[0-9][^0-9]%', string + 't') - PATINDEX('%[0-9]%', 
                    string) + 1) AS Number
FROM @temp

Answer 4

Please try:

declare @var nvarchar(max)='Balance1000sheet'

SELECT LEFT(Val,PATINDEX('%[^0-9]%', Val+'a')-1) from(
    SELECT SUBSTRING(@var, PATINDEX('%[0-9]%', @var), LEN(@var)) Val
)x

Answer 5

Getting only numbers from a string can be done in a one-liner. Try this :

SUBSTRING('your-string-here', PATINDEX('%[0-9]%', 'your-string-here'), LEN('your-string-here'))

NB: Only works for the first int in the string, ex: abc123vfg34 returns 123.

Answer 6

I found this approach works about 3x faster than the top voted answer. Create the following function, dbo.GetNumbers:

CREATE FUNCTION dbo.GetNumbers(@String VARCHAR(8000))
RETURNS VARCHAR(8000)
AS
BEGIN;
    WITH
        Numbers
    AS (
        --Step 1.
        --Get a column of numbers to represent
        --every character position in the @String.
        SELECT 1 AS Number
        UNION ALL
        SELECT Number + 1
        FROM Numbers
        WHERE Number < LEN(@String)
        )
        ,Characters
    AS (
        SELECT Character
        FROM Numbers
        CROSS APPLY (
                --Step 2.
                --Use the column of numbers generated above
                --to tell substring which character to extract.
                SELECT SUBSTRING(@String, Number, 1) AS Character
            ) AS c
        )
    --Step 3.
    --Pattern match to return only numbers from the CTE
    --and use STRING_AGG to rebuild it into a single string.
    SELECT @String = STRING_AGG(Character,'')
    FROM Characters
    WHERE Character LIKE '[0-9]'

    --allows going past the default maximum of 100 loops in the CTE
    OPTION (MAXRECURSION 8000) 

    RETURN @String
END
GO

Testing

Testing for purpose:

SELECT dbo.GetNumbers(InputString) AS Numbers
FROM ( VALUES
         ('003Preliminary Examination Plan') --output: 003
        ,('Coordination005')                 --output: 005
        ,('Balance1000sheet')                --output: 1000
        ,('(111) 222-3333')                  --output: 1112223333
        ,('1.38hello@f00.b4r#\-6')           --output: 1380046
    ) testData(InputString)

Testing for performance: Start off setting up the test data...

--Add table to hold test data
CREATE TABLE dbo.NumTest (String VARCHAR(8000)) 

--Make an 8000 character string with mix of numbers and letters
DECLARE @Num VARCHAR(8000) = REPLICATE('12tf56se',800)

--Add this to the test table 500 times
DECLARE @n INT = 0
WHILE @n < 500
BEGIN
    INSERT INTO dbo.NumTest VALUES (@Num)
    SET @n = @n +1
END

Now testing the dbo.GetNumbers function:

SELECT dbo.GetNumbers(NumTest.String) AS Numbers
FROM dbo.NumTest -- Time to complete: 1 min 7s

Then testing the UDF from the top voted answer on the same data.

SELECT dbo.udf_GetNumeric(NumTest.String)
FROM dbo.NumTest -- Time to complete: 3 mins 12s

Inspiration for dbo.GetNumbers

Decimals

If you need it to handle decimals, you can use either of the following approaches, I found no noticeable performance differences between them.

• change '[0-9]' to '[0-9.]'
• change Character LIKE '[0-9]' to ISNUMERIC(Character) = 1 (SQL treats a single decimal point as "numeric")

Bonus

You can easily adapt this to differing requirements by swapping out WHERE Character LIKE '[0-9]' with the following options:

• WHERE Letter LIKE '[a-zA-Z]' --Get only letters
• WHERE Letter LIKE '[0-9a-zA-Z]' --Remove non-alphanumeric
• WHERE Letter LIKE '[^0-9a-zA-Z]' --Get only non-alphanumeric

Answer 7

A solution for SQL Server 2017 and later, using TRANSLATE:

DECLARE @T table (string varchar(50) NOT NULL);

INSERT @T 
    (string)
VALUES 
    ('003Preliminary Examination Plan'),
    ('Coordination005'),
    ('Balance1000sheet');

SELECT 
    result =
        REPLACE(
            TRANSLATE(
                T.string COLLATE Latin1_General_CI_AI, 
                'abcdefghijklmnopqrstuvwxyz', 
                SPACE(26)), 
            SPACE(1), 
            SPACE(0))
FROM @T AS T;

Output:

result
003
005
1000

The code works by:

1. Replacing characters a-z (ignoring case & accents) with a space
2. Replacing spaces with an empty string.

The string supplied to TRANSLATE can be expanded to include additional characters.

Answer 8

With the previous queries I get these results:

'AAAA1234BBBB3333' >>>> Output: 1234

'-çã+0!\aº1234' >>>> Output: 0

The code below returns All numeric chars:

1st output: 12343333

2nd output: 01234

declare @StringAlphaNum varchar(255)
declare @Character varchar
declare @SizeStringAlfaNumerica int
declare @CountCharacter int

set @StringAlphaNum = 'AAAA1234BBBB3333'
set @SizeStringAlfaNumerica = len(@StringAlphaNum)
set @CountCharacter = 1

while isnumeric(@StringAlphaNum) = 0
begin
    while @CountCharacter < @SizeStringAlfaNumerica
        begin
            if substring(@StringAlphaNum,@CountCharacter,1) not like '[0-9]%'
            begin
                set @Character = substring(@StringAlphaNum,@CountCharacter,1)
                set @StringAlphaNum = replace(@StringAlphaNum, @Character, '')
            end
    set @CountCharacter = @CountCharacter + 1
    end
    set @CountCharacter = 0
end
select @StringAlphaNum

Answer 9

declare @puvodni nvarchar(20)
set @puvodni = N'abc1d8e8ttr987avc'

WHILE PATINDEX('%[^0-9]%', @puvodni) > 0 SET @puvodni = REPLACE(@puvodni, SUBSTRING(@puvodni, PATINDEX('%[^0-9]%', @puvodni), 1), '' ) 

SELECT @puvodni

Answer 10

This UDF will work for all types of strings:

CREATE FUNCTION udf_getNumbersFromString (@string varchar(max))
RETURNS varchar(max)
AS
BEGIN
    WHILE  @String like '%[^0-9]%'
    SET    @String = REPLACE(@String, SUBSTRING(@String, PATINDEX('%[^0-9]%', @String), 1), '')
    RETURN @String
END

Answer 11

I did not have rights to create functions but had text like

["blahblah012345679"]

And needed to extract the numbers out of the middle

Note this assumes the numbers are grouped together and not at the start and end of the string.

select substring(column_name,patindex('%[0-9]%', column_name),patindex('%[0-9][^0-9]%', column_name)-patindex('%[0-9]%', column_name)+1)
from table name

Answer 12

Just a little modification to @Epsicron 's answer

SELECT SUBSTRING(string, PATINDEX('%[0-9]%', string), PATINDEX('%[0-9][^0-9]%', string + 't') - PATINDEX('%[0-9]%', 
                    string) + 1) AS Number
FROM (values ('003Preliminary Examination Plan'),
    ('Coordination005'),
    ('Balance1000sheet')) as a(string)

no need for a temporary variable

Answer 13

Although this is an old thread its the first in google search, I came up with a different answer than what came before. This will allow you to pass your criteria for what to keep within a string, whatever that criteria might be. You can put it in a function to call over and over again if you want.

declare @String VARCHAR(MAX) = '-123.  a    456-78(90)'
declare @MatchExpression VARCHAR(255) = '%[0-9]%'
declare @return varchar(max)

WHILE PatIndex(@MatchExpression, @String) > 0
    begin
    set @return = CONCAT(@return, SUBSTRING(@string,patindex(@matchexpression, @string),1))
    SET @String = Stuff(@String, PatIndex(@MatchExpression, @String), 1, '')
    end
select (@return)

Answer 14

Firstly find out the number's starting length then reverse the string to find out the first position again(which will give you end position of number from the end). Now if you deduct 1 from both number and deduct it from string whole length you'll get only number length. Now get the number using SUBSTRING

declare @fieldName nvarchar(100)='AAAA1221.121BBBB'

declare @lenSt int=(select PATINDEX('%[0-9]%', @fieldName)-1)
declare @lenEnd int=(select PATINDEX('%[0-9]%', REVERSE(@fieldName))-1)

select SUBSTRING(@fieldName, PATINDEX('%[0-9]%', @fieldName), (LEN(@fieldName) - @lenSt -@lenEnd))

Answer 15

T-SQL function to read all the integers from text and return the one at the indicated index, starting from left or right, also using a starting search term (optional):

create or alter function dbo.udf_number_from_text(
    @text nvarchar(max),
    @search_term nvarchar(1000) = N'',
    @number_position tinyint = 1,
    @rtl bit = 0
) returns int
as
    begin
        declare @result int = 0;
        declare @search_term_index int = 0;

        if @text is null or len(@text) = 0 goto exit_label;
        set @text = trim(@text);
        if len(@text) = len(@search_term) goto exit_label;

        if len(@search_term) > 0
            begin
                set @search_term_index = charindex(@search_term, @text);
                if @search_term_index = 0 goto exit_label;
            end;

        if @search_term_index > 0
            if @rtl = 0
                set @text = trim(right(@text, len(@text) - @search_term_index - len(@search_term) + 1));
            else
                set @text = trim(left(@text, @search_term_index - 1));
        if len(@text) = 0 goto exit_label;

        declare @patt_number nvarchar(10) = '%[0-9]%';
        declare @patt_not_number nvarchar(10) = '%[^0-9]%';
        declare @number_start int = 1;
        declare @number_end int;
        declare @found_numbers table (id int identity(1,1), val int);

        while @number_start > 0
        begin
            set @number_start = patindex(@patt_number, @text);
            if @number_start > 0
                begin
                    if @number_start = len(@text)
                        begin
                            insert into @found_numbers(val)
                            select cast(substring(@text, @number_start, 1) as int);

                            break;
                        end;
                    else
                        begin
                            set @text = right(@text, len(@text) - @number_start + 1);
                            set @number_end = patindex(@patt_not_number, @text);

                            if @number_end = 0
                                begin
                                    insert into @found_numbers(val)
                                    select cast(@text as int);

                                    break;
                                end;
                            else
                                begin
                                    insert into @found_numbers(val)
                                    select cast(left(@text, @number_end - 1) as int);

                                    if @number_end = len(@text)
                                        break;
                                    else
                                        begin
                                            set @text = trim(right(@text, len(@text) - @number_end));
                                            if len(@text) = 0 break;
                                        end;
                                end;
                        end;
                end;
        end;

        if @rtl = 0
            select @result = coalesce(a.val, 0)
            from (select row_number() over (order by m.id asc) as c_row, m.val
                    from @found_numbers as m) as a
            where a.c_row = @number_position;
        else
            select @result = coalesce(a.val, 0)
            from (select row_number() over (order by m.id desc) as c_row, m.val
                    from @found_numbers as m) as a
            where a.c_row = @number_position;


        exit_label:
            return @result;
    end;

Example:

select dbo.udf_number_from text(N'Text text 10 text, 25 term', N'term',2,1);

returns 10;

Answer 16

This is one of the simplest and easiest one. This will work on the entire String for multiple occurences as well.

CREATE FUNCTION dbo.fn_GetNumbers(@strInput NVARCHAR(500))
RETURNS NVARCHAR(500)
AS
BEGIN
DECLARE @strOut NVARCHAR(500) = '', @intCounter INT = 1
WHILE @intCounter <= LEN(@strInput) 
BEGIN
    SELECT @strOut = @strOut + CASE WHEN SUBSTRING(@strInput, @intCounter, 1) LIKE '[0-9]' THEN SUBSTRING(@strInput, @intCounter, 1) ELSE '' END    
    SET @intCounter = @intCounter + 1 

END
RETURN @strOut
END 

Answer 17

Following a solution using a single common table expression (CTE).

DECLARE @s AS TABLE (id int PRIMARY KEY, value nvarchar(max));

INSERT INTO @s
VALUES
    (1, N'003Preliminary Examination Plan'),
    (2, N'Coordination005'),
    (3, N'Balance1000sheet');

SELECT * FROM @s ORDER BY id;

WITH t AS (
    SELECT
        id,
        1 AS i,
        SUBSTRING(value, 1, 1) AS c
    FROM
        @s
    WHERE
        LEN(value) > 0

    UNION ALL

    SELECT
        t.id,
        t.i + 1 AS i,
        SUBSTRING(s.value, t.i + 1, 1) AS c
    FROM
        t
        JOIN @s AS s ON t.id = s.id
    WHERE
        t.i < LEN(s.value)
)
SELECT
    id,
    STRING_AGG(c, N'') WITHIN GROUP (ORDER BY i ASC) AS value
FROM
    t
WHERE
    c LIKE '[0-9]'
GROUP BY
    id
ORDER BY
    id;

Answer 18

DECLARE @index NVARCHAR(20);
SET @index = 'abd565klaf12';

WHILE PATINDEX('%[0-9]%', @index) != 0
BEGIN
    SET @index = REPLACE(@index, SUBSTRING(@index, PATINDEX('%[0-9]%', @index), 1), '');
END

SELECT @index;

One can replace [0-9] with [a-z] if numbers only are wanted with desired castings using the CAST function.

Answer 19

If we use the User Define Function, the query speed will be greatly reduced. This code extracts the number from the string....

        SELECT 
           Reverse(substring(Reverse(rtrim(ltrim( substring([FieldName] , patindex('%[0-9]%', [FieldName] ) , len([FieldName]) )))) , patindex('%[0-9]%', Reverse(rtrim(ltrim( substring([FieldName] , patindex('%[0-9]%', [FieldName] ) , len([FieldName]) )))) ), len(Reverse(rtrim(ltrim( substring([FieldName] , patindex('%[0-9]%', [FieldName] ) , len([FieldName]) ))))) )) NumberValue
        FROM dbo.TableName

Answer 20

CREATE OR REPLACE FUNCTION count_letters_and_numbers(input_string TEXT)
RETURNS TABLE (letters INT, numbers INT) AS $$
BEGIN
    RETURN QUERY SELECT 
        sum(CASE WHEN input_string ~ '[A-Za-z]' THEN 1 ELSE 0 END) as letters,
        sum(CASE WHEN input_string ~ '[0-9]' THEN 1 ELSE 0 END) as numbers
    FROM unnest(string_to_array(input_string, '')) as input_string;
END;
$$ LANGUAGE plpgsql;

Answer 21

I got this solution:

with SL as (select 0 as STEP,@VALUE as VAL
  union all select STEP+1,replace(VAL,SUBSTRING(VAL,patindex('%[^0-9]%',VAL),1),'') from SL where patindex('%[^0-9]%',VAL)>0)
  select top 1 VAL from SL order by STEP desc

It can work if string's length is fitted in substring function (8000 for varchar or 4000 for nvarchar) and contains not more than 100 different non-numeric characters — upper and lower case is counted or not according to your collation settings.

Answer 22

For the hell of it...

This solution is different to all earlier solutions, viz:

• There is no need to create a function
• There is no need to use pattern matching
• There is no need for a temporary table
• This solution uses a recursive common table expression (CTE)

But first - note the question does not specify where such strings are stored. In my solution below, I create a CTE as a quick and dirty way to put these strings into some kind of "source table".

Note also - this solution uses a recursive common table expression (CTE) - so don't get confused by the usage of two CTEs here. The first is simply to make the data avaliable to the solution - but it is only the second CTE that is required in order to solve this problem. You can adapt the code to make this second CTE query your existing table, view, etc.

Lastly - my coding is verbose, trying to use column and CTE names that explain what is going on and you might be able to simplify this solution a little. I've added in a few pseudo phone numbers with some (expected and atypical, as the case may be) formatting for the fun of it.

with SOURCE_TABLE as (
    select '003Preliminary Examination Plan' as numberString
    union all select 'Coordination005' as numberString
    union all select 'Balance1000sheet' as numberString
    union all select '1300 456 678' as numberString
    union all select '(012) 995 8322  ' as numberString
    union all select '073263 6122,' as numberString
),
FIRST_CHAR_PROCESSED as (
    select
        len(numberString) as currentStringLength,
        isNull(cast(try_cast(replace(left(numberString, 1),' ','z') as tinyint) as nvarchar),'') as firstCharAsNumeric,
        cast(isNull(cast(try_cast(nullIf(left(numberString, 1),'') as tinyint) as nvarchar),'') as nvarchar(4000)) as newString,
        cast(substring(numberString,2,len(numberString)) as nvarchar) as remainingString
    from SOURCE_TABLE
    union all
    select
        len(remainingString) as currentStringLength,
        cast(try_cast(replace(left(remainingString, 1),' ','z') as tinyint) as nvarchar) as firstCharAsNumeric,
        cast(isNull(newString,'') as nvarchar(3999)) + isNull(cast(try_cast(nullIf(left(remainingString, 1),'') as tinyint) as nvarchar(1)),'') as newString,
        substring(remainingString,2,len(remainingString)) as remainingString
    from FIRST_CHAR_PROCESSED fcp2
    where fcp2.currentStringLength > 1
)
select 
    newString
    ,* -- comment this out when required
from FIRST_CHAR_PROCESSED 
where currentStringLength = 1

So what's going on here?

Basically in our CTE we are selecting the first character and using try_cast (see docs) to cast it to a tinyint (which is a large enough data type for a single-digit numeral). Note that the type-casting rules in SQL Server say that an empty string (or a space, for that matter) will resolve to zero, so the nullif is added to force spaces and empty strings to resolve to null (see discussion) (otherwise our result would include a zero character any time a space is encountered in the source data).

The CTE also returns everything after the first character - and that becomes the input to our recursive call on the CTE; in other words: now let's process the next character.

Lastly, the field newString in the CTE is generated (in the second SELECT) via concatenation. With recursive CTEs the data type must match between the two SELECT statements for any given column - including the column size. Because we know we are adding (at most) a single character, we are casting that character to nvarchar(1) and we are casting the newString (so far) as nvarchar(3999). Concatenated, the result will be nvarchar(4000) - which matches the type casting we carry out in the first SELECT.

If you run this query and exclude the WHERE clause, you'll get a sense of what's going on - but the rows may be in a strange order. (You won't necessarily see all rows relating to a single input value grouped together - but you should still be able to follow).

Hope it's an interesting option that may help a few people wanting a strictly expression-based solution.

Answer 23

In Oracle

You can get what you want using this:

SUBSTR('ABCD1234EFGH',REGEXP_INSTR ('ABCD1234EFGH', '[[:digit:]]'),REGEXP_COUNT ('ABCD1234EFGH', '[[:digit:]]'))

Sample Query:

SELECT SUBSTR('003Preliminary Examination Plan  ',REGEXP_INSTR ('003Preliminary Examination Plan  ', '[[:digit:]]'),REGEXP_COUNT ('003Preliminary Examination Plan  ', '[[:digit:]]')) SAMPLE1,
SUBSTR('Coordination005',REGEXP_INSTR ('Coordination005', '[[:digit:]]'),REGEXP_COUNT ('Coordination005', '[[:digit:]]')) SAMPLE2,
SUBSTR('Balance1000sheet',REGEXP_INSTR ('Balance1000sheet', '[[:digit:]]'),REGEXP_COUNT ('Balance1000sheet', '[[:digit:]]')) SAMPLE3 FROM DUAL

Answer 24

If you are using Postgres and you have data like '2000 - some sample text' then try substring and position combination, otherwise if in your scenario there is no delimiter, you need to write regex:

SUBSTRING(Column_name from 0 for POSITION('-' in column_name) - 1) as 
number_column_name